Quadratic Functions and Equations - DDTwo

Quadratic Functions and Equations

What is Quadratic Function? Equation?

How to Solve Quadratic Function/ Equation?

Friday, January 31, 2020

Quadratic Function and Equation

A quadratic function is in the standard form y = ax2 + bx + c or f(x) = ax2 + bx + c

A quadratic equation is a quadratic function equated to zero

The standard quadratic equation form is ax2 + bx + c = 0 where a, b, and c are numbers with a 0.

Solving Quadratic Equation

Quadratic Equation can be solve by

Taking the square root

Factoring

Completing the square

Quadratic formula

Graphing calculator



Only Quadratic Equation of the form

ax2 + c = 0

Example:

Which of the following quadratic equation can be solved by taking the square root?

can be solved by taking the square root.

B.) 3x2 – 9x = 0A.) 2x2 + 8 = 0

C.) x2 + 4x – 5 = 0 D.) none of these


Example: Solve by taking the square root:

1.) 2x2 – 8 = 0 2.) (x – 3)2 + 8 = 44

2x2 = 8

2 2

4

x2 = 4

x2 =

x = 2

x = 2 Or x = -2

(x – 3)2 = 36

(x – 3)2 = 36

x – 3 = 6

x = 6 + 3 Or x = -6 + 3

x = 9 Or x = -3


Example: Solve by taking the square root:

3.) 2x2 + 2 = 0 4.) (x – 3)2 + 8 = -28

2x2 = -2

2 2

-1

x2 = -1

x2 =

x = i

x = i Or x = -i

(x – 3)2 = -36

(x – 3)2 = -36

x – 3 = 6i

x = 3 + 6i Or x = 3 – 6i

Numbers with i is called an imaginary number

Real and Imaginary together are called Complex number


Try this! : Solve by taking the square root:

5.) x2 – 8 = 0 6.) (x + 1)2 + 8 = 58

x2 = 8

8x2 =

x = 2

x = 2

Or x = -2

(x + 1)2 = 50

(x + 1)2 = 50

x + 1 = 5

x = -1 + 5

Or x = -1 – 5

2

2

2

2

2

2

Solving Q. E. by Factoring

Example: Factor and use the zero property to solve the Q.E.

1. x² + 3x − 10 = 0

= 0( x ) ( x ) 2 and -5 -3

-2 and 5 3

1 and -10 -9

-1 and 10 9

- 2 + 5

Factors of -10 Sum of Factors

Using zero property

x – 2 = 0 or x + 5 = 0

x = 2 or x = -5




2. x² − 5x + 6 = 0

= 0( x ) ( x ) 2 and 3 5

1 and 6 7

-2 and -3 -5

-1 and -6 -7

- 2 - 3

Factors of 6 Sum of Factors

Using zero property

x – 2 = 0 or x – 3 = 0

x = 2 or x = 3



3. x² − 2x − 3 = 0

= 0( x ) ( x ) 1 and -3 -2

-1 and 3 2+ 1 - 3

Factors of -3 Sum of Factors

Using zero property

x + 1 = 0 or x – 3 = 0

x = -1 or x = 3


Example: Box Method Factoring and the zero property to solve the Q.E.

F of -18 Sum

-9, 2 -7

9,-2 7

-6,3 -3

6,-3 3

1,-18 -17

-1,18 17

1. 3x2 + 7x – 6 = 0

1st: (3)(-6) = -18

2nd:3rd: Box Method

3x2

-6

9x

-2x

x 3

3x

-2

(x + 3)(3x – 2) = 0

GCF

x + 3 = 03x – 2 = 0+2 +2

3x = 23 3

x 23

=

x =-3



F of -168 Sum

-21, 8 -13

21,-8 13

-12,14 2

12,-14 -2

28,-6 22

-28,6 17

2. 8x2 + 22x – 21 = 0

1st: (8)(-21) = -168

2nd:3rd: Box Method

8x2

-21

28x

-6x

2x 7

4x

-3

(2x + 7) (4x – 3) = 0

GCF

Example: Box Method Factoring and the zero property to solve the Q.E.

4x – 3 = 0+3 +3

4x = 34 4

x 34

=

2x + 7 = 0- 7 -7

2x = -72 2

x -72

=


Example: Solve by factoring

1.) 6x2 + 7x – 5 = 0

2.) x2 – 6x = 27

(2x - 1)(3x + 5) = 0

(2x - 1) = 0 Or (3x + 5) = 0

x2 – 6x – 27 = 0

(x + 3)(x – 9) = 0

(x + 3) = 0 Or (x – 9) = 0

x = -3 Or x = 9

x = 1/2 Or x = -5/3


Solving by Completing the Square

Recall: Perfect Square Trinomials

Examples

x2 + 6x + 9

x2 - 10x + 25

x2 + 12x + 36

= (x + 3)(x + 3)

= (x + 3)2

= (x + 6)(x + 6)

= (x – 5)(x – 5)

= (x – 5)2

= (x + 6)2


Creating a Perfect Square Trinomial

In the following perfect square trinomial, the constant term is missing. X2 + 14x + ____

Find the constant term by squaring half the coefficient of the linear term (the number beside x).

(14/2)2

X2 + 14x + 49

Creating a Perfect Square Trinomial

Create a perfect square trinomial and factor

x2 + 20x + ___

x2 - 4x + ___

x2 + 5x + ___

100

4

25/4

= (x + 10)2

= (x – 2)2

= (x + 5/2)2


Solving by Completing the Square



Solve by Completing the Square Example 1

Step 1: Move quadratic term, and linear term to left side and the constant term to right side of the equation

x2 + 8x – 20 = 0

+ 20 +20

x2 + 8x = 20Step 2: Find the number that completes the square on the left side and add to both sides.of equation

x2 + 8x = 20+ 16 +16

Step 3: Factor the left side of the equation and simplify the right side of the equation

(x + 4)2 = 36

Solve by Completing the Square

2( 4) 36x

( 4) 6x


Step 4: Solve by taking the square root

4 6

4 6 an

d 4 6

10 and 2 x=

x

x x

x

Solve by Completing the Square Example 2

22 7 12 0x x

22 7 12x x


Step 1: Move quadratic term, and linear term to left side and the constant term to right side of the equation

Step 2:

Find the term that completes the square on the left side of the equation. Add that term to both sides.

2

2

2

2 7

2

2 2 2

7 12

7

2

=-12 +

6

x x

x x

xx

21 7 7 49

( ) then square it, 2 62 4 4 1

7

2 49 49

16 1

76

2 6x x


The quadratic coefficient must be equal to 1 before you complete the square, so you must divide all terms by the quadratic coefficient first.

Step 3:

Factor the perfect square trinomial on the left side of the equation. Simplify the right side of the equation.

2

2

2

76

2

7 96 49

4 16 16

7 47

4

49 49

16 1

16

6x x

x

x


27 47( )

4 16x

7 47( )

4 4

7 47

4 4

7 47

4

x

ix

ix


Step 4: Solve by taking the square root

2

2

2

2

2

1. 2 63 0

2. 8 84 0

3. 5 24 0

4. 7 13 0

5. 3 5 6 0

x x

x x

x x

x x

x x

Try the following examples. Do your work on your paper and then check your answers.

1. 9,7

2.(6, 14)

3. 3,8

7 34.

2

5 475.

6

i

i


Solve by Completing the Square


Solving by Quadratic Formula

ax2 + bx + c = 0

ax2 + bx = -c

a

cx

a

bx2

2

2

2

22

a4

b

a

c

a4

b x

a

bx

The quadratic formula is derived by completing the square using the standard quadratic equation:

2

22

a4

ac4b

a2

bx


Solving by Quadratic Formula

a2

ac4bbx

2

The Quadratic Formula is

The standard quadratic equation form is ax2 + bx + c = 0 where a, b, and c are numbers with a 0.

All quadratic equation can be solved by using the quadratic formula



Solve by quadratic formula

1.) 2x2 + 4x = 5 2x2 + 4x – 5 = 0

a2

ac4bbx

2

a=2 ,b=4, and c=-5

)2(2

)5)(2(4)4(4 2 x

4

40164x

4

564x

4

1424x

2

142x



2.) x2 + 13 = 4x x2 – 4x + 13 = 0

a2

ac4bbx

2

a=1 ,b=-4, and c=13

)1(2

)13)(1(4)4()4(x

2

2

52164x

2

364x

2

i64x

i32x

Changing equation in Standard

Change the equation in standard quadratic form and identify a,b, and c then solve using Quadratic Formula

1.) 2x2 + 8 = 0

4.) (x – 3)2 + 8 = 44

2.) 3x2 = 9x

3.) x2 + 4x = 5

a=2 ,b=0, and c=8

3x2 – 9x = 0

2x2 + 8 = 0

a=3 ,b=-9, and c=0

x2 + 4x – 5 = 0

a=1 ,b=4 and c=-5

x2 – 6x – 27 = 0

a=1 ,b=-6, and c=-27

ax2 + bx + c = 0Standard form



1.) x2 – 3x – 2 = 0

2.) 2x2 + 13 = 8x

3.) 3x2 – 3 = 4x

Find Quadratic Equation

Finding a Quadratic Function or Equation given its Roots, Zeros or Solution.

Example:

1. Find a Quadratic function/ equation whose zeros are x = 1 and x = -3

Solution:

f(x) =

(x – 1) (x + 3)

f(x) = x2 + 3x - 1x - 3

f(x) = x2 + 2x - 3 or x2 + 2x – 3 = 0

Find Quadratic Equation

Finding a Quadratic Function or Equation given its Roots, Zeros or Solution.

Try it yourself:

2. Find a Quadratic function/ equation whose zeros are x = 2 and x = -1

Solution:

f(x) =

(x – 2) (x + 1)

f(x) = x2 + 1x - 2x - 2

f(x) = x2 - 1x - 2 or x2 - 1x – 2 = 0

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Quadratic Functions and Equations - DDTwo